# Why is it trivial for compact Riemann surfaces to have a countable topology?

In "Lectures on Riemann surfaces" by Otto Forster, before the proof of Rado's theorem which asserts that every Riemann surface has a countable topology, the author commented "Clearly this is trivial for compact Riemann surfaces". This is not clear at all for me. Could someone help me understand this ?

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Riemann surfaces do not have countable topologies. Do you mean second countable? –  Chris Eagle Dec 23 '11 at 13:18
@Aki: You mean "has a countable base for its topology", a.k.a. is second-countable. This is a much weaker condition than "has a countable topology". –  Zev Chonoles Dec 23 '11 at 13:18
What is "countable topology"? Does it mean that it has a topology which is first or second countable? –  Paul Dec 23 '11 at 13:20
Thank you @ChrisEagle and ZevChonoles for quick answers. The above is the direct quotation from the book. I understand it as second countable or "has a countable base for its topology" as you suspect. –  Aki Dec 23 '11 at 13:27
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## 1 Answer

Such a surface is a union of finitely many open disks. Each disk has a countable basis consisting of balls of rational radius around points with rational coordinates. The union of the bases for each disk is a countable basis for the whole space.

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Thank you ! Now it is clear for me too. –  Aki Dec 23 '11 at 13:49
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